AP50 Fall 2016
Problem Set 5
due Thursday, November 17th in class
1. Estimate the damping coefficient for a shock absorber on a midsize car.
Getting Started:
As a vehicle drives over a bump in the road, we want the body of the car
to remain level. This is accomplished by having a spring attached to each
wheel which can compress or extend in the vertical direction. As we know
from Hooke’s law, a deflected spring has a restoring force that will
immediately pull/push it back to its equilibrium position. Without energy
dissipation, the spring will continue to oscillate back and forth around the
equilibrium position, which defeats the goal of keeping the car level. This
problem is addressed by adding damping components to the car. These components
quickly reduce the spring’s oscillation magnitude and keep the ride smooth. Therefore,
we can model this system as a damped simple harmonic oscillator with a spring and
damper at each wheel (Ch. 15.8). http://www.thecartech.com/subjects/auto_eng2/
INTRODUCTION_TO_SUSPENSION.htm
Devise a Plan:
As described above, let’s model the car suspension system as a mass with a spring and a
damper at each wheel. This system is modeled by the equation:
The total spring constant shown in the sketch above, k, is a sum of the
four springs at each wheel which we assume all to have equal spring
constants individually, k = k1 + k2 + k3 + k4 = 4*k1. See sketch at right.
Similarly, for the dampers we assume total damping constant, b, with b
= b1 + b2 + b3 + b4 = 4*b1. We will use the mass of a midsize car, which
we will estimate using the fact that an angry mob can flip a car.
Damping causes the oscillations to reduce in amplitude: light damping causes the
oscillations to decay slowly, whereas heavier damping causes the oscillations to decay
quickly. We can estimate the amount of damping by estimating how quickly the
oscillation amplitude decreases. We can then estimate τ, the time constant, using
equation 15.39. Finally, we will use the estimated mass and the relationship τ=m/b to
find the overall damping.
Execute Plan:
We first estimate the mass of a midsize car. To estimate this mass, we recall incidents of
angry mobs flipping a car. This requires lifting one side over the other, which requires
lifting half the weight of the car. We estimate that 10-15 people could fit along the side
of a car, and each could lift around 100kg. This means they can flip a car of 1000-1500kg.
Since mobs would be more successful with the lighter cars, we will assume a typical
midsize car is on the heavy side of this range, so we will use a mass of 1500kg.
Next, we estimate how the amplitude changes with each cycle. After 1 full cycle, the
amplitude is around 10% of the original disturbance. We estimate the cycle takes about
1s, so we use equation 15.39 to find τ:
AP50 Fall 2016
so and [s].
Next ,we use τ = m/b to determine b. We estimated m = 1500[kg], so b = 1500/0.22 [kg/
s] ≈ 6900 [kg/s].
This is the total damping for the car. The damping at each wheel accounts for ¼ of the
total damping, so b1 = b/4 = 6900/4 = 1700[kg/s] at each wheel.
Evaluate Result:
Our estimate hinges on two quantities: the mass of a car and the damping time constant.
To check the mass estimate, we use the fact that a loaded car handles quite differently
from an empty car. This implies that the mass of the additional people must not be
negligible compared to the mass of the car. A loaded car has 4 additional people, each
with mass around 70kg, so 4*70kg = 280kg is not negligible compared to the mass of the
car. Therefore, the car should be within an order of magnitude of 280kg. We estimated
1500kg, roughly 5x more than the weight of the people, so our mass is likely at least the
right order of magnitude.
Next ,we check the time constant. A time constant of 0.22 means that about 4 time
constants will pass within a period. However, this is a time constant for the energy decay:
E = E0*exp(-t/τ). Since energy scales with the square of the amplitude, the amplitude has
a time constant that is twice as long: xmax = A*exp(-t/2τ). Thus, having 4 time constants
pass in one period means that the amplitude should be A*exp(-4τ/2τ) = A*(e-1)2≈A/9. This
means that the amplitude is reduced 9-fold in one period, which is close to the 10-fold
reduction we assumed. This confirms that we did not make a mistake computing the time
constant.
Here is another way to estimate the damping. For ease of calculations, we see from Figure
15.37 that we want b > 0.5*mω since we want our oscillations to die out ideally before we
even see one full oscillation.
We estimated the period T=1s, so ω = 2*π/T ≈ 6.28 [s−1]. Thus, b> 0.5mω =
0.5*1500[kg]*6.28[s−1] = 4700[kg/s]. This is a lower bound on the overall damping. This
value is lower than what we computed above, but it is also the same order of magnitude,
justifying our answer.
AP50 Fall 2016
Just for fun – a neat animation http://www.acoustics.salford.ac.uk/feschools/waves/
shm4.php that we can play with to see how the quality factor Q can be damped by shock
absorbers in a vehicle and what the Q factor means in terms of a resonance of the springs
in the suspension system as a function of vehicle speed.
AP50 Fall 2016
2. Swinging Baseball Bat: A hole is drilled through the narrow end of a 0.960-kg
baseball bat, and the bat is hung on a nail so that it can swing freely (see
figure). The bat is 0.860 m long, the hole is drilled 0.0300 m from the end,
and the center of mass is located 0.670 m from the end. If the period of
oscillation is 1.85 s, what is the bat’s rotational inertia (a) about the pivot
and (b) about the center of mass? (c) If the hole had been drilled 0.200 m
from the end, what would the period be?
Getting Started:
We have a baseball bat that rotates around a pivot (nail).
As discussed in Ch 15.7 (Restoring Torques), here we have a
pendulum which is a type of rotational oscillator. In this
case, the restoring force is not caused by an elastic force
such as a spring but rather by the force of Earth’s gravity
on the baseball ball which causes a torque on the bat. The
torque creates a pendulum-like motion and the bat can
swing with some period while hanging from the nail.
We have drawn an extended free body diagram of the bat
since this problem involves torques.
Devise a Plan:
We can split the force shown on the extended free body
diagram above into parallel and perpendicular components
and find that the torque caused by the force of gravity on
the bat about the pivot is:
where ℓcm is the lever arm for this force. For small angles of rotation we approximate
and therefore find that the torque is linearly proportional to the rotational
displacement.
Torque is equal to rotational inertia times angular acceleration (rotational version of F =
ma):
Substituting this into eq. 15.31 we can get an equation for the second derivative of the
rotational position in terms of the rotational position.
This equation (15.32) is just like our translational equation for simple harmonic motion
(eq. 15.21), so its solution has the same form.
AP50 Fall 2016
We know that the angular frequency of oscillation for a simple harmonic oscillator (like a
mass on a spring) is given as in 15.22; similarly, the angular frequency for the pendulum
we have described above can be given as:
For the rest of the problem we will apply this equation with slightly different setups.
When the axis around which we want to rotate changes, we will need to use the parallel
axis theorem from chapter 11 to find the rotational inertias.
Execute Plan:
(part a) Rearrange equation 15.33 to solve for the rotational inertia, I.
(part b) Use the parallel axis theorem (Chapter 11.6) and the rotational inertia from part
(a) to find the rotational inertia around the center of mass. The two parallel axes are the
axis going through the center of mass into the page and the axis going through the pivot
into a page. The distance between these axes is 0.640m.
(part c) If the hole were 0.200 m from the end instead of 0.0300 m, the distance from the
pivot to the center of mass would now be 0.470m. Use the parallel axis theorem again to
find the new rotational inertia.
AP50 Fall 2016
Now that we have the new rotational inertia, we can set equation 15.33 for angular
frequency of oscillation, ω equal to 2π/T and rearrange to solve for the period, T.
Evaluate Result:
When we put the nail in the bat closer and closer to the center, we are effectively making
the rotating mass section of the bat shorter so we would expect a shorter period. We see
this from part a to part c where the period goes from 1.62 s to 1.4 s as the nail gets closer
to the mass.
We also see that the moment of inertia is smallest at the center of mass, bigger when the
pivot is 0.47m away from the center of mass, and bigger still when the pivot is 0.64m
away from the center of mass, as expected.
From “Example 15.6 The simple pendulum” in the textbook, the period of a simple
pendulum is given by:
Using this equation (meaning that we assume the mass is a ball concentrated at the end of
the full length of the bat instead of at the center of mass) we find the maximum period is
1.86s. This is the upper limit for a simple pendulum of length 0.860 m.
This is somewhat close to the period of 1.62s given in the problem in part a when the nail
hole is only 0.030 m from the end. It is longer which makes sense because here we ignore
that the mass is not centered on the end of the bat. We can think of the “effective
length” of our simple pendulum from the pivot to the mass as being a little shorter than
what we actually calculate above which would make the period a little shorter (agreeing
with what we calculated directly in part a).
AP50 Fall 2016
3. Musician: A flute is a tube open at both ends. It will emit a musical note if a standing wave is
excited in the air column inside the tube. (a) The lowest musical note that a musician can play on
this flute is C (261.7 Hz). What must be the length of the tube? Assume that the air column is
vibrating in its fundamental mode. (b) In order to produce higher musical notes, a musician opens
valves along the tube. Since the holes in these valves are large, an open valve has the same effect
as shortening the tube. The musician opens one valve to play C#, two valves to play D, etc.
Calculate the successive spacing between the valves of a flute for one complete octave. (The
actual spacing used on flutes differs slightly from the results of this simple theoretical evaluation
because the mouth cavity of the flutist affects the frequency.) Note that going up an octave
corresponds to doubling the note frequency and each octave is divided into 12 notes whose
frequencies are chosen to sound equally-spaced. This is accomplished by setting the ratio between
the frequencies of adjacent notes to the same value, which works because the musical interval we
hear between two notes corresponds to the ratio between their frequencies. The value of this
ratio, r, is determined by the fact that going up 12 notes gives an octave, which has double the
frequency of the original note, so multiplying the ratio 12 times needs to double the frequency: f0
x r12 = 2 x f0, and hence the ratio is r=21/12. Thus, the frequency of the C# is 21/12 times higher than
that of the C; the D is 21/12 times higher than the C# or 22/12 times higher than the C, etc.
Getting Started:
We are modeling a flute as an open cylinder of length, L as shown in the sketch below. In our
simplified model, we assume that all valves are closed to play the fundamental C as shown in the
sketch below.
http://newt.phys.unsw.edu.au/jw/fluteacoustics.html
In this model, as we open the first valve along the tube, the cylinder is now effectively shorter
meaning the next note (C#/Db) will have a shorter wavelength (and higher frequency). We open the
second valve and the next note (D) will have an even shorter wavelength, etc. until we reach C
again another octave higher.
Devise a Plan:
First we will find the length of the tube (with all valves closed) that
corresponds to the longest wavelength (lowest frequency) that could be
played on this instrument. This is the fundamental frequency, and we are
told that it is the C at 261.7 Hz. This lowest frequency corresponds to
the longest wavelength that forms a standing wave in the flute. As shown
in the sketch above, the longest wavelength is twice as long as the flute
itself:.
www.musictheoryacademy.com
We know the velocity because the pressure wave is travelling at the speed of sound through air.
We are also given the frequency.
AP50 Fall 2016
Next, we will use an equal-tempered scale to calculate the frequency of each note in the octave
We will do this for an entire octave, as easily visualized on the piano keys shown above. We know
that the frequency of each note is 21/12 times higher than the previous note. We then use the
equation above to find the length of the tube required for each frequency; this is where the valve
is positioned for that note.
Execute Plan:
(part a)
We calculate the length of the flute using the frequency of the fundamental C = 261.7 Hz.
(part b)
We create a spreadsheet to calculate the frequencies and lengths of the remaining notes in the
octave. The frequencies are each consecutively multiplied by and then the formula above is used
again to calculate the new tube length for each given frequency. They are subtracted sequentially
to find the spacing between holes (ΔL). We use the speed of sound in dry air at 20 °C as 343 m/s
(wikipedia)
Note
Frequency (Hz)
Length (m)
ΔL (mm)
C
261.7
0.655
0
C#/Db
277.3
0.619
37
D
293.7
0.584
35
D#/Eb
311.2
0.551
33
E
329.7
0.520
31
F
349.3
0.491
29
F#/Gb
370.1
0.463
28
G
392.1
0.437
26
G#/Ab
415.4
0.413
25
A
440.1
0.390
23
A#/Bb
466.3
0.368
22
B
494.0
0.347
21
C
523.4
0.328
19
Evaluate Result:
The values that we calculated for the frequencies of the notes match those found in the following
website, starting from note C4 to C5:
http://www.phy.mtu.edu/~suits/notefreqs.html
A note one octave higher should have twice the frequency, as we confirmed for C = 261.7 Hz and
523.4 Hz.
AP50 Fall 2016
AP50 Fall 2016
4. Whistling footballs: Two students standing 54.9m apart are tossing around footballs designed to
whistle while they are airborne. (a) When student A throws her football toward student B at 16.0
m/s, it emits a frequency of 680 Hz. What frequency is heard by student B? (b) When student B
throws his football toward student A at 15.0 m/s, it emits a frequency 670 Hz. What frequency is
heard by student A? (c) If the two students throw the footballs toward one another simultaneously,
what is the beat frequency?
AP50 Fall 2016
AP50 Fall 2016
Below is the link to hyperphysics on beat frequencies:
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html
5.
Whale of a Time: On a whale-watching expedition, your underwater sound detector picks up a
whale sound that has an intensity of 9.0 µW/m2. (a) What is the intensity level β of this sound? (b)
If your detector tells you that you are 2.3 km from the whale, what is the power of the emitted
signal? (c) How many whales, emitting sound at the same intensity I = 9.0 µW/m2, are needed to
produce an intensity level of 100dB at your location?
AP50 Fall 2016
AP50 Fall 2016